Triplet whose sum in range (1,2) - arrays

Question

Given n positive real numbers in an array, find whether there exists a triplet
among this set such that, the sum of the triplet is in the range (1,
2). Do it in linear time and constant space.
the array is not ordered.
numbers are positive
numbers are real numbers
Any help would be greatly appreciated. Thanks.

Answer 1

The trick is to figure out a way to categorize the possible solutions and come up with a linear-time constant-space solution for each.
Consider the three ranges X = (0,2/3), Y = [2/3,1], Z = (1,2). At most one value can come from Z (if two values came from Z, then the sum would exceed 1+1=2). Similarly, at least one value must come from X. Suppose there were 3 values a <= b <= c so that 1 <= a+b+c <= 2 . Then, consider what possible classes of solutions are feasible:
A) `a \in X, b \in X, C \in X`
B) `a \in X, b \in X, C \in Y`
C) `a \in X, b \in X, C \in Z`
D) `a \in X, b \in Y, C \in Y`
E) `a \in X, b \in Y, C \in Z`
So how can we test each case?
Case A is incredibly easy to test: the sum is guaranteed to be below 2, so we just need to test the largest sum (largest 3 elements in X) exceeds 1.
Case C is incredibly easy to test: since the sum is guaranteed to be above 1, we only need to check if the sum is below 2. So in order to do that, we just need to test the smallest 2 values in X and the smallest value in Z
Cases D and E are similar to C (since the sum must be at least 4/3 > 1, choose the smallest possible values in each class).
Case B is the only tricky case. 0 < a+b < 4/3 and 2/3 <= c <= 1. To handle case B, we consider these intervals : X1 = (0, 1/2), X2 = [1/2 2/3), Y = [2/3, 1].
This results in following three valid cases :
B1. a in X1, b in X2, c in Y
B2. a in X1, b in X1, c in Y
B3. a in X2, b in X2, c in Y
Case B1 & B3 : Sum of three numbers is always greater than 1 so we take minimum values and check if it is smaller than 2 or not.
Case B2 : Sum of three numbers is always less than 2, so we take maximum sum and check if is greater than 1 or not.
So to summarize, the tests are:
|X| >= 3 and Xmax(1) + Xmax(2) + Xmax(3) >= 1
|X| >= 2, |Z| >= 1, and Xmin(1)+Xmin(2)+Zmin(1) <= 2
|X| >= 1, |Y| >= 2, and Xmin(1)+Ymin(1)+Ymin(2) <= 2
|X| >= 1, |Y| >= 1, |Z| >= 1, and Xmin(1)+Ymin(1)+Zmin(1) <= 2
|X| >= 2, |Y| >= 1, and Xmax(1) + Xmax(2) + Ymin(1) < 2
|X| >= 2, |Y| >= 1, and Xmin(1) + Xmin(2) + Ymax(1) > 1)
Each test can be performed in linear time and constant space (you only need to find Xmax(1), Xmax(2), Xmax(3), Xmin(1), Xmin(2), Ymin(1), Ymin(2), Ymax(1), Zmin(1), all of which can be found in one pass even if the data is not sorted)

Answer 2

So, you have an array of double data types of length n. Intialize three variables a,b and c as first 3 values of array.Now,iterate from i = 3 to n and check the following: 1)Check if sum falls in (1, 2),if it does then return true. 2)If not, then check if sum is greater than 2,if so, then replace MAX(a,b,c) to current element arr[i]. 3)Otherwise sum must be less than 1 then replace MIN(a,b,c) to current element arr[i].And finally after coming out of loop check once again for last triplet if sum falls in (1,2) then return true,otherwise return false.
enter code here
double a=arr[0], b=arr[1], c=arr[2];
for(int i=3 ; i<n ; i++){
// check if sum fall in (1, 2)
if(a+b+c > 1 && a+b+c < 2){
return 1;
}
// if not, then check is sum greater than 2
// if so, then replece MAX(a,b,c) to new number
else if(a+b+c > 2){
if(a>b && a>c){
a = arr[i];
}
else if(b>a && b>c){
b = arr[i];
}
else if(c>a && c>b){
c = arr[i];
}
}
// else then sum must be less than 1
// then replace MIN(a,b,c) to new number
else{
if(a<b && a<c){
a = arr[i];
}
else if(b<a && b<c){
b = arr[i];
}
else if(c<a && c<b){
c = arr[i];
}
}
}
// check for last a, b, c triplet
if(a+b+c > 1 && a+b+c < 2){
return 1;
}
else{
return 0;
}

Answer 3

Question asked is similar to this InterviewBit question. I've made some changes to the solution mentioned below to match exactly what you asked.
There are three conditions:
First, if the sum is between 1 and 2.
Second, if the sum is greater than 2, then the larger element of the three will be replaced with the next element and the process is repeated.
Third, if the sum is less than 1, then the smaller element of the three will be replaced with the next element and same process is repeated.
int Solution::solve(vector<float> &A) {
if(A.size()<3) return 0;
float a = A[0];
float b = A[1];
float c = A[2];
for(int i=3;i<A.size();++i){
if(a+b+c>1 && a+b+c<2)
return 1;
float temp = A[i];
if(a+b+c>=2){
if(a>b && a>c)
a = temp;
else if(b>c && b>a)
b = temp;
else
c = temp;
}
else{
if(a<b && a<c)
a = temp;
else if(b<c && b<a)
b = temp;
else
c = temp;
}
}
if(a+b+c>1 && a+b+c<2) return 1;
return 0;
}
Same question can also be solved using two pointer technique. In this first we'll have to sort the array. But, It's complexity will be more than O(logn).
int Solution::solve(vector<float> &A) {
int n = A.size();
if(n<3) return 0;
sort(A.begin(), A.end());
int l = 0, r = n-1;
while(l<r-1){
float s = A[l]+A[l+1]+A[r];
if(s>=2)
r = r-1;
else if(s<1)
l = l+1;
else return 1;
}
return 0;
}

Answer 4

This problem can easily be solved in linear runtime time using the sliding window sum approach. In this case, we will use a window of size 3. This is also called the 'moving sum algorithm'.
Algorithm Below
1> Prepare the window of size 3 with the first 3 elements
2> IF (array.len <= 3): CHECK IF window-sum is in the range (1,2), then RETURN accordingly
3> FOR i = 3 UPTO (array.len-1)
3.1> SORT the window (3log3 = constant time operation)
3.2> IF window-sum is in the range (1,2): RETURN 1 or TRUE
3.3> ELSE IF window-sum < 1: Replace the smallest element in the window (window[0]) with array[i]
3.4> ELSE IF window-sum > 2: Replace the largest element in the window (window[2]) with array[i]
4> Outside the loop, check the FINAL window sum and RETURN accordingly.
Access the Python code here. Star the repository, please!

Answer 5

Java code for the solution given by #soul Ec.
we need to modify the case B.
let say our numbers are a+b+c
there are three ranges
x1 x2 y
(0,1/2) (1/2,2/3) (2/3,1)
we have 4 possibilities
1. x1 + x1 +y
2. x2 + x2 +y
3. x1 + x2 +y
4 x2 + x1 +y
here case 3 and 4 are identical as sum of it will be same. so we have 3 cases only.
1. x1 + x1 + y it is always <2 ( do x1max+x1max+ymax <2 to verify)
so we have to check if x1max(1)+x1max(2)+ymax(1) > 1
2. x2 + x2 + y it is always >1 ( do x2min+x2min+ymin >1 to verify)
so we have to check if x2min(1)+x2min(2)+ymin(1) <=2
3. x1 + x2 + y it is always >1 (do x1min+x2min+ymin >1 to verify)
so we have to check if x1min(1)+x2min(1)+ymin(1)<=2
public static int solve(ArrayList<String> A) {
double d[]= new double[A.size()];
for(int i=0;i<A.size();i++) {
d[i]= Double.parseDouble(A.get(i));
}
double range1 = 0;
double range2 = (double) 2/3;
double range3 = 1;
double range4 = 2;
double range02 =(double) 1/2;
// min and max in range (0,2/3)
double min1= Double.MAX_VALUE;
double min2=Double.MAX_VALUE;
double min3=Double.MAX_VALUE;
double max1= Double.MIN_VALUE;
double max2=Double.MIN_VALUE;
double max3=Double.MIN_VALUE;
// min and max in range (2/3,1)
double miny1= Double.MAX_VALUE;
double miny2=Double.MAX_VALUE;
double miny3=Double.MAX_VALUE;
double maxy1= Double.MIN_VALUE;
double maxy2=Double.MIN_VALUE;
double maxy3=Double.MIN_VALUE;
// min and max in range (1,2)
double minz1= Double.MAX_VALUE;
double minz2=Double.MAX_VALUE;
double minz3=Double.MAX_VALUE;
double maxz1= Double.MIN_VALUE;
double maxz2=Double.MIN_VALUE;
double maxz3=Double.MIN_VALUE;
// min and max in range (0,1/2)
double minxx1= Double.MAX_VALUE;
double minxx2=Double.MAX_VALUE;
double minxx3=Double.MAX_VALUE;
double maxx1= Double.MIN_VALUE;
double maxx2=Double.MIN_VALUE;
double maxx3=Double.MIN_VALUE;
// min and max in range (1/2,2/3)
double minyy1= Double.MAX_VALUE;
double minyy2=Double.MAX_VALUE;
double minyy3=Double.MAX_VALUE;
double maxyy1= Double.MIN_VALUE;
double maxyy2=Double.MIN_VALUE;
double maxyy3=Double.MIN_VALUE;
for (int i = 0; i < d.length; i++) {
if (d[i] >= range1 && d[i] < range02) {
if (d[i] < minxx3) {
minxx1=minxx2;
minxx2=minxx3;
minxx3 = d[i];
} else if (d[i] > minxx3 && d[i] < minxx2) {
minxx1=minxx2;
minxx2 = d[i];
} else if (d[i] > minxx3 && d[i] > minxx2 && d[i] < minxx1) {
minxx1 = d[i];
}
if (d[i] > maxx3) {
maxx1=maxx2;
maxx2=maxx3;
maxx3 = d[i];
} else if (d[i] < maxx3 && d[i] > maxx2) {
maxx1=maxx2;
maxx2 = d[i];
} else if (d[i] < maxx3 && d[i] < maxx2 && d[i] > maxx1) {
maxx1 = d[i];
}
}
if (d[i] >= range02 && d[i] < range2) {
if (d[i] < minyy3) {
minyy1=minyy2;
minyy2=minyy3;
minyy3 = d[i];
} else if (d[i] > minyy3 && d[i] < minyy2) {
minyy1=minyy2;
minyy2 = d[i];
} else if (d[i] > minyy3 && d[i] > minyy2 && d[i] < minyy1) {
minyy1 = d[i];
}
if (d[i] > maxyy3) {
maxyy1=maxyy2;
maxyy2=maxyy3;
maxyy3 = d[i];
} else if (d[i] < maxyy3 && d[i] > maxyy2) {
maxyy1=maxyy2;
maxyy2 = d[i];
} else if (d[i] < maxyy3 && d[i] < maxyy2 && d[i] > maxyy1) {
maxyy1 = d[i];
}
}
if (d[i] >= range1 && d[i] < range2) {
if (d[i] < min3) {
min1=min2;
min2=min3;
min3 = d[i];
} else if (d[i] > min3 && d[i] < min2) {
min1=min2;
min2 = d[i];
} else if (d[i] > min3 && d[i] > min2 && d[i] < min1) {
min1 = d[i];
}
if (d[i] > max3) {
max1=max2;
max2=max3;
max3 = d[i];
} else if (d[i] < max3 && d[i] > max2) {
max1=max2;
max2 = d[i];
} else if (d[i] < max3 && d[i] < max2 && d[i] > max1) {
max1 = d[i];
}
}
if (d[i] >= range2 && d[i] < range3) {
if (d[i] < miny3) {
miny1=miny2;
miny2=miny3;
miny3 = d[i];
} else if (d[i] > miny3 && d[i] < miny2) {
miny1=miny2;
miny2 = d[i];
} else if (d[i] > miny3 && d[i] > miny2 && d[i] < miny1) {
miny1 = d[i];
}
if (d[i] > maxy3) {
maxy1=maxy2;
maxy2=maxy3;
maxy3 = d[i];
} else if (d[i] < maxy3 && d[i] > maxy2) {
maxy1=maxy2;
maxy2 = d[i];
} else if (d[i] < maxy3 && d[i] < maxy2 && d[i] > maxy1) {
maxy1 = d[i];
}
}
if (d[i] >= range3 && d[i] <= range4) {
if (d[i] < minz3) {
minz1=minz2;
minz2=minz3;
minz3 = d[i];
} else if (d[i] > minz3 && d[i] < minz2) {
minz1=minz2;
minz2 = d[i];
} else if (d[i] > minz3 && d[i] > minz2 && d[i] < minz1) {
minz1 = d[i];
}
if (d[i] > maxz3) {
maxz1=maxz2;
maxz2=maxz3;
maxz3 = d[i];
} else if (d[i] < maxz3 && d[i] > maxz2) {
maxz1=maxz2;
maxz2 = d[i];
} else if (d[i] < maxz3 && d[i] < maxz2 && d[i] > maxz1) {
maxz1 = d[i];
}
}
}
if(max1+max2+max3>=1 && max1!=Double.MIN_VALUE && max2!=Double.MIN_VALUE && max3!=Double.MIN_VALUE)
return 1;
if(min3+min2+minz3<=2 && min3!=Double.MAX_VALUE && min2!=Double.MAX_VALUE && minz3!=Double.MAX_VALUE )
return 1;
if(min3+miny3+miny2<=2 && min3!=Double.MAX_VALUE && miny3!=Double.MAX_VALUE && miny2!=Double.MAX_VALUE)
return 1;
if(min3+miny3+minz3<=2 && min3!=Double.MAX_VALUE && miny3!=Double.MAX_VALUE && minz3!=Double.MAX_VALUE)
return 1;
if(maxx3+maxx2+maxy3>1 && maxx3!=Double.MIN_VALUE && maxx2!=Double.MIN_VALUE && maxy3!=Double.MIN_VALUE) {
return 1;
}
if(minyy3+minyy2+miny3<=2 && minyy3!=Double.MAX_VALUE && minyy2!=Double.MAX_VALUE && miny3!=Double.MAX_VALUE) {
return 1;
}
if(minxx3+minyy3+miny3<=2 && minxx3!=Double.MAX_VALUE && minyy3!=Double.MAX_VALUE && miny3!=Double.MAX_VALUE) {
return 1;
}
return 0;
}

Answer 6

the solution is in c++(interviewbbit solution)
int Solution::solve(vector<string> &arr) {
int n=arr.size(),i;
vector<float>v;
for(i=0;i<n;i++)
{
v.push_back(stof(arr[i]));
}
float a=v[0],b=v[1],c=v[2];
float mx=0;
for(i=3;i<n;i++)
{
if(a+b+c<2 && a+b+c>1)
return 1;
else if(a+b+c>2)
{
if(a>b && a>c)
a=v[i];
else if(b>a && b>c)
b=v[i];
else
c=v[i];
}
else
{
if(a<b && a<c)
a=v[i];
else if(b<a && b<c)
b=v[i];
else
c=v[i];
}
}
if(a+b+c>1 && a+b+c<2)
return 1;
else
return 0;
}

Answer 7

We can easily do it in O(n) , we just need to find three minimum positive real numbers ,which we can find in one iteration and in the end if their summation lies in between (1,2) then return 1 else return 0.

Answer 8

The problem in its whole as stated is undecidable. This is due to the fact that for any two real numbers a and b it cannot be decided if a > b holds (also see this answer of me). But you have to do at least one comparison of a real number against an integer value to solve that problem. Comparing against an integer doesn't make the problem easier since you could have a real number which is 2,00...001 where there is an arbitrary number of zeros between the digits 2 and 1 which you don't know beforehand. Storing such real numbers (probably not every one, but many of them) in the main memory of a computer is not a big problem for such specific ones, since they can be represented by an approximation algorithm.
for further information I suggest reading Complexity Theory of Real Functions